Department of Chemistry (Japanese)2 exp ν/k B T []()hν/k B −1 2 (4.1.12) 9CEF =8. 9CEF =8. KY,...
Transcript of Department of Chemistry (Japanese)2 exp ν/k B T []()hν/k B −1 2 (4.1.12) 9CEF =8. 9CEF =8. KY,...
19
656.5 nm 486.3
nm 434.2 nm 410.3 nm
J.J. 1885 λ
1
λ= RH
1
22−1
n2⎛
⎝ ⎜
⎞
⎠ ⎟ (4.1.1)
RH 109678 cm-1
n = 3 4 5 6 n = 7 397.1 nm n =
8 389.0 nm (4.1.1)
20 (4.1.1)
1
λ= RH
1
l2−1
n2⎛
⎝ ⎜
⎞
⎠ ⎟ (4.1.2)
l = 1 1906 l = 3 1908
l = 4 1922 l = 2
(4.1.2)
n
l
ρ(ν) =8πν 2
c 3hν
exp hν /kBT( ) −1 (4.1.3)2
h T
kB 1
R R = NAkB (4.1.3)
A. 1905 ν hν
ν
hν hν/c
1
W 4.2
K
2 exp(x) ex x
θ
(mV )2 =hν ic
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
+hν oc
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
− 2hν ic
⎛
⎝ ⎜
⎞
⎠ ⎟ hν oc
⎛
⎝ ⎜
⎞
⎠ ⎟ cosθ
=h2
c 2ν i2 + ν o
2 − 2ν iν o cosθ( ) (4.1.5)
=h2
c 2ν i −ν o( )2 + 2ν iν o 1− cosθ( )[ ]
m2V 2 = 2h2
c 2ν iν o 1− cosθ( ) (4.1.6)
1
2mV 2 = h ν i −ν o( ) (4.1.7)
m ν i −ν o( ) = h
c 2ν iν o 1− cosθ( ) (4.1.8)
V λ
c = νλ (4.1.9)
(4.1.8)
Δλ = λo − λi =h
mc1− cosθ( ) (4.1.10)
h m c
C
N
C = 3R ×N
NA
(4.1.11)
4.4 + 1000 K
3 1819
L.E. 1871
1 mol 3R
4.4 1905 W.H.
0
A. 1907
ν hν
C =N
NA
⎛
⎝ ⎜
⎞
⎠ ⎟ kB
hνkBT
⎛
⎝ ⎜
⎞
⎠ ⎟
2exp hν /kBT( )
exp hν /kBT( ) −1[ ]2 (4.1.12)
r
F = –Ar/|r|3 (4.2.1)
A
ke2 =e2
4πε0 (4.2.2)
e ε0
Ae = 2.3•10-28 N m-2 G m M
GmM Ag = 2.7•10-39 N m-2 Ag << Ae
r 4
mV 2
r= k
e2
r2 (4.2.3)
V (4.2.3)
1
2mV 2 + −k
e2
r
⎛
⎝ ⎜
⎞
⎠ ⎟ = −
1
2ke2
r (4.2.4)
r
10-11 s
A
N.H.D.
1913
= h/2π
n
mVr = n (4.2.5)
(4.2.3) V r
r = n2
2
kme2 (4.2.6)
(4.2.6) n
a0 =
2
kme2 (4.2.7)
a0
52.9 pm
(4.2.6)
r = n2a0 (4.2.8)
r (4.2.4)
En = −1
2n2ke2
a0 (4.2.9)
n
Eh =
ke2
a0=k 2me4
2 (4.2.10)
Eh
n (4.2.9)
i f i > f
ΔE = Ei − E f = −1
ni2 −
1
n f2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ 1
2Eh (4.2.11)
ν λ hν
1
n f2 −
1
ni2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ 1
2Eh = hν = h
c
λ (4.2.12)
1
λ=
1
n f2 −
1
ni2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ 1
2hcEh (4.2.13)
m mM/(m + M)
1
λ=Eh
2hc
1
1+ (m /M)1
n f2 −
1
ni2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ (4.2.14)
(4.1.2)
RH =Eh
2hc
1
1+ (m /M) (4.2.15)
1923 λ
p
λ =h
p (4.3.1)
(4.2.5) mV
2πr = nλ (4.3.2)
50 kg 6 km h-1 80 N s
h/p ≈ (6.6•10-34 J s)/(80 N s) ≈ 8•10-36 m
100 V 0.1 nm
1925 W.M.
4.5 F(x) F(x ± ct)
x t = 0 F(x)
F(x − ct) c F(x + ct)
c 4.5 ψ(x, t) = F(x ± ct)
B
∂∂xψ(x, t) =
∂X∂x
dF(X)
dX=dF(X)
dX
∂∂tψ(x, t) =
∂X∂t
dF(X)
dX= ±c
dF(X)
dX (4.3.3)
∂2
∂t 2ψ(x, t) = c 2
d2F(X)
dX 2
ψ(x, t)
∂2
∂x 2ψ(x, t) −
1
c 2∂2
∂t 2ψ(x,t) = 0 (4.3.4)
5 ψ(x, t) = F(x ± ct)
F(x − ct) F(x + ct)
a F(x − ct) + bF(x + ct) (4.3.4)
5
∂2
∂x 2+∂2
∂y 2+∂2
∂z2⎛
⎝ ⎜
⎞
⎠ ⎟ ψ(x,y,z,t) −
1
c 2∂2
∂t 2ψ(x,y,z,t) = 0 (4.3.5)
(4.3.5)
∇
∇ =∂∂x,∂∂y,∂∂z
⎛
⎝ ⎜
⎞
⎠ ⎟ (4.3.6)
∇2 =∇ •∇ =∂2
∂x 2+∂2
∂y 2+∂2
∂z2 (4.3.7)
∇2ψ −1
c 2∂2
∂t 2ψ = 0 (4.3.8)
A ν λ x
ψ = Asin 2πx
λ−νt
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥ (4.3.9)
(4.3.4) c λ ν
c = νλ (4.3.10)
x0 x0
∂∂tψ = −2πνAcos 2π
x0λ−νt
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥ (4.3.11)
ν 2A2
ψ =ψ0(x,y,z)exp(i2πνt) (4.3.12)
B
exp(iθ) = eiθ = cosθ + isinθ (4.3.13)
∂2
∂t 2ψ = (i2πν)2ψ0(x,y,z)exp(i2πνt) = −(2πν )
2ψ (4.3.14)
(4.3.8)
∇2ψ +4π2ν 2
c 2ψ = 0 (4.3.15)
∇2ψ +4π2
λ2ψ = 0 (4.3.16)
E U(x, y, z)
E =1
2mv 2 +U =
1
2mp2 +U (4.3.17)
p2 = 2m[E – U(x, y, z)] p λ
(4.3.1) (4.3.16) λ
∇2ψ +8mπ2
h2[E −U(x,y,z)]ψ = 0 (4.3.18)
−
2
2m∇2 +U(x,y,z)
⎡
⎣ ⎢
⎤
⎦ ⎥ ψ = Eψ (4.3.19)
(4.3.18) (4.3.19) E. 1926
–ke2/r